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Math Help - Equation system excercise

  1. #1
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    Equation system excercise

    Solve for every a the equationsystem:

    <br />
x+y+az=3<br />
    <br />
2x-y+3x=4<br />
    <br />
x+ay+z=3<br />

    I put the system in a matrix and did elimination until:

    1 2 a 3
    0 -3 (3-2a) -2
    0 (a-1) (1-a) 0

    which gives me a = (y-z)/(y-z)=1

    Is this correct, I'm not very used to this kind of exercises with unknown coefficients so I'm rather unsure at this. Am I going in the right direction? If not I'd be grateful for any help!
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  2. #2
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    No one can help me?
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  3. #3
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    Quote Originally Posted by Boneleg View Post
    Solve for every a the equationsystem:


    x+y+az=3

    2x-y+3z=4 <<<<<<<<<< typo?

    x+ay+z=3


    I put the system in a matrix and did elimination until:

    1 2 a 3
    0 -3 (3-2a) -2
    0 (a-1) (1-a) 0

    which gives me a = (y-z)/(y-z)=1

    Is this correct, I'm not very used to this kind of exercises with unknown coefficients so I'm rather unsure at this. Am I going in the right direction? If not I'd be grateful for any help!
    1. I assume that there is a typo at the indicated line(?)

    2. In my opinion you have to solve this system for (x, y, z) with respect to a.

    3. I used determinants and I got (x, y, z)=\left(2-\dfrac1a\ ,\ \dfrac1a\ ,\ \dfrac1a \right)~,~a \neq 0
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